POJ 1328 Radar Installation

                                  

Radar Installation
                                      Time Limit:1000MS     Memory Limit:10000KB     64bit
IO Format:%I64d & %I64u

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance,
so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of
the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 



Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed
by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2Case 2: 1

思路：转化成区间问题，如果区间有相交叉的部分，说明有一个雷达就能覆盖，
如果，两区间不相交，说明一个雷达覆盖不了，只能再用一个雷达。这就成了
求区间上不相交的小区间个数了

#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;

struct point
{
double left;               //因为会出现小数所以定义为double型
double right;
} a[1001];

bool cmp( const point x, const point y)
{

return x.right<y.right;
}

int main()
{
int n,d,t=1;
while(scanf("%d%d",&n,&d)!=EOF)
{
if(n==0&&d==0)
break;
int i,count=0;
double x,y,cur;
int flag=1;
for(i=0; i<n; i++)
{
scanf("%lf%lf",&x,&y);
printf("%lf %lf\n",x,y);
if(d-y<0)             //如果小岛所在的纵坐标大于雷达所覆盖的范围
flag=0;           //计为0
else if(flag)
{
a[i].left=x-sqrt(d*d-y*y);  //计算雷达离某小岛的最远距离的左右坐标
a[i].right=x+sqrt(d*d-y*y);
}
}
if(!flag)     //如果为0，说明不能完全覆盖，输出 -1
{
printf("Case %d: -1\n",t++);
continue;
}
sort (a,a+n,cmp);  //对右坐标从小到大排序
cur=a[0].right;
for(i=1; i<n; i++)
{
if(cur<a[i].left)   //如果左边的坐标（开始点）大于上一个区间的结束点坐标
{
count++;
cur=a[i].right;
}
}
printf("Case %d:\n",t++);
printf("%d\n",count);
}
return 0;
}